Problem: Simplify and expand the following expression: $ \dfrac{1}{q + 7}+ \dfrac{5}{q - 8}+ \dfrac{3q}{q^2 - q - 56} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{3q}{q^2 - q - 56} = \dfrac{3q}{(q + 7)(q - 8)}$ Now we have: $ \dfrac{1}{q + 7}+ \dfrac{5}{q - 8}+ \dfrac{3q}{(q + 7)(q - 8)} $ The least common multiple of the denominators is: $ (q + 7)(q - 8)$ In order to get the first term over $(q + 7)(q - 8)$ , multiply by $\dfrac{q - 8}{q - 8}$ $ \dfrac{1}{q + 7} \times \dfrac{q - 8}{q - 8} = \dfrac{q - 8}{(q + 7)(q - 8)} $ In order to get the second term over $(q + 7)(q - 8)$ , multiply by $\dfrac{q + 7}{q + 7}$ $ \dfrac{5}{q - 8} \times \dfrac{q + 7}{q + 7} = \dfrac{5(q + 7)}{(q + 7)(q - 8)} $ Now we have: $ \dfrac{q - 8}{(q + 7)(q - 8)} + \dfrac{5(q + 7)}{(q + 7)(q - 8)} + \dfrac{3q}{(q + 7)(q - 8)} $ $ = \dfrac{ q - 8 + 5(q + 7) + 3q} {(q + 7)(q - 8)} $ Expand: $ = \dfrac{q - 8 + 5q + 35 + 3q}{q^2 - q - 56} $ $ = \dfrac{9q + 27}{q^2 - q - 56}$